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What is the value of $\psi (1)$ ? If we take the definition in terms of derivative of Gamma function, we get $\psi (1) = \dfrac{\Gamma'(1)}{\Gamma(1)} = -\gamma$. But, if we consider the series representation i.e, $\psi (a) = - \displaystyle \sum_{n=0}^{\infty} \dfrac{1}{n+a}$, then $\psi(1) = - \zeta(1)$ which diverges.

Here, $\psi (x)$ is the digamma function

Why are there contradictory answers?

Any help will be appreciated.
Thanks.

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You've misread the series representation. It should be \begin{align} \psi(n)&=-\gamma+\sum_{k=1}^{n-1}\frac1k\\ &=H_{n-1}-\gamma \end{align} where $H_n$ denotes the $n$th Harmonic number. Note that this is only defined for natural numbers $n$ (where the sum from $1$ to $0$ is the so-called "empty sum", which is equal to $0$. Also note that $H_0=0$). So actually you've misread two things, since also $\psi(1)\neq\gamma$, but $\psi(1)=-\gamma$.

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  • $\begingroup$ But according to wikipedia, the series representation is true. See en.wikipedia.org/wiki/Digamma_function#Regularization $\endgroup$ – user313478 Feb 18 '16 at 9:33
  • $\begingroup$ I'm not sure what they mean by "regularization of divergent integrals", so I unfortunately cannot answer that question. Should I delete my answer? $\endgroup$ – vrugtehagel Feb 18 '16 at 9:57
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    $\begingroup$ @FlamingNinja: You misread what Wikipedia says. That article speaks about attributing finite values to divergent series $($ which is what regularization means $),$ and $\psi(a)$ is s very logical choice, in certain contexts, for the sum in question. $\endgroup$ – Lucian Feb 18 '16 at 18:41
  • $\begingroup$ @Lucian Oh ok. Thanks! $\endgroup$ – user313478 Mar 17 '16 at 5:35

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